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I have solved this problem. I wish to find out if my solution is correct. I am a bit confuse with the second part of question (b).

Problem: Let S be the surface of a solid $R$ , which lies inside the cylinder: $$x^2+y^2=16$$ and between the plane

where $x=0$ and $z=5$

There is also defined a vector field F by: $$\begin{align}F(x,y)=(-x^3i-y^3j+3z^2k)\end{align}$$

(a) Calculate : $$\iint_{T} F.\hat n\mathrm dS$$

with T = {(x,y,5)$\in$ $\mathbb{R^3}$|$x^2+y^2\le16$}

(b) Calculate DivF and $$\iint_{S} F.\hat n\mathrm dS$$

with n the outward pointing unit normal.

(c) Calculate: $$\iint_{V} F.\hat n\mathrm dS$$

with V = {(x,y,z)$\in$ $\mathbb{R^3}$|$x^2+y^2=16$ and $0\le$z$\le$5} and the unit normal $\hat n$ points out of the solid $R$

Solution: (a) On the top surface of the Cylinder z = 5, $\hat n$ = $\hat k$ $$F.\hat n = [-x^3i +-y^3j +3z^2k].[k]=3z^2 $$ $$\iint_{S} F.\hat n \mathrm dS = \iint_{S} 3z^2\mathrm dS$$ $$\iint_{S} F.\hat n \mathrm dS = \iint_{S} 3(5^2)\mathrm dS$$ $$\iint_{S} F.\hat n \mathrm dS = 75\iint_{S}\mathrm dS$$ The area enclosed by the circle is $\pi$$r^2$ = 16$\pi$ since the radius of the circle is 4. Therefore $$\iint_{S} F.\hat n \mathrm dS = 75(16\pi) = 1200\pi $$

(b) $$DivF = \nabla.F = [i\frac{\partial }{\partial x}+ j\frac{\partial }{\partial y} +k\frac{\partial }{\partial z}].[-x^3i +-y^3j +3z^2k]= -3x^2-3y^2 +6z$$ $$DivF = \nabla.F = -3(x^2+y^2-2z)$$ $$\iint_{S} F.\hat n \mathrm dS =\iiint_{V} \operatorname{div} F dV -\iint_{S_1} F.\hat n \mathrm dS$$ $$ \iint_{S_1} F.\hat n \mathrm dS = 0 $$ since z=0, then $$ \iint_{S} F.\hat n \mathrm dS = \iiint_{V} \operatorname{div} F dV $$

How do I get $\hat n$ in this case?

(c) From my understanding, I have to use Divergence Theorem here $$\iint_{V} F.\hat n\mathrm dS = \iiint_{V} \nabla.F \mathrm dV $$ $$\iint_{V} F.\hat n\mathrm dS = \iiint_{V} -3(x^2+y^2-2z) \mathrm dV $$ $$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} (x^2+y^2-2z) \mathrm dV $$

Using Cylindrical coordinates $$\iint_{V} F.\hat n\mathrm dS = -3\iiint_{V} [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta $$ $$\iint_{V} F.\hat n\mathrm dS = -3\int_0^{2\pi} \int_0^4 \int_0^5 [(r^2\cos^2\theta+r^2\sin^2\theta-2z)] \mathrm rdzdrd\theta $$ $$\iint_{V} F.\hat n\mathrm dS = 432\pi $$

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Use the Divergence Theorem for the whole thing. This converts the surface integral to a volume integral:

$$ \iint_T F\cdot \hat{n} dS = \iiint_V (\nabla \cdot F) dV$$

This is much more doable to finish and it gets rid of $\hat{n}$.

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HINT:

As you have already found $\operatorname{div} {F}=\nabla.F = -3(x^2+y^2-2z)$.

Now, for part (b) imagine the region $S$ , you will find that it is not bounded below. So, if you wish to apply Gass'Divergence theorem here , it will be wrong** . To apply it first you have to bound the region. Let us bound it below by plane $z=0$.

Now,

$\iint_{S \cup S_1} F.\hat n \mathrm dS =\iiint_{V} \operatorname{div} F dV$

where,

$S$ : the surface as stated in the question.

$S_1$: the surface of plane $z=0$ inside $(x^2+y^2=16 , x=0 , z=5)${actually, it is semi disk in positive side of $x$ -axis}

$V$ : the volume inside the region $S \cup S_1$

$ \hat{n}$: outward drawn normal to the surface $S \cup S_1$

Therefore,

$\iint_{S} F.\hat n \mathrm dS =\iiint_{V} \operatorname{div} F dV -\iint_{S_1} F.\hat n \mathrm dS$

Note: **Why? see here Mathematical Statement https://en.wikipedia.org/wiki/Divergence_theorem

Let me know if you are able to solve this part now.

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  • $\begingroup$ If z=0, then the flux of surface S1 will be zero $\endgroup$ – Soso Apr 4 '18 at 9:05

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